Problem 31 - A Textbook of Differential Calculus

Question

If \(y=\tan x \tan 2 x \tan 3 x\), then \(\frac{d y}{d x}\) has the value equal to (a) \(3 \sec ^{2} 3 x \tan x \tan 2 x+\sec ^{2} x \tan 2 x \tan 3 x+2 \sec ^{2} 2 x \tan 3 x \tan x\) (b) \(2 y(\operatorname{cosec} 2 x+2 \operatorname{cosec} 4 x+3 \operatorname{cosec} 6 x)\) (c) \(3 \sec ^{2} 3 x-2 \sec ^{2} 2 x-\sec ^{2} x\) (d) \(\sec ^{2} x+2 \sec ^{2} 2 x+3 \sec ^{2} 3 x\)

Step-by-Step Solution

Answer

The correct answer is (a).

Step 1: Understanding the Problem

We're given the function \( y = \tan x \tan 2x \tan 3x \), and we need to find its derivative, \( \frac{dy}{dx} \). The options provided represent different forms of derivative expressions or identities, and we'll need to differentiate the expression correctly to identify the right one.

Step 2: Apply the Product Rule

The function is a product of three tangent functions: \( y = u \cdot v \cdot w \), where \( u = \tan x \), \( v = \tan 2x \), and \( w = \tan 3x \). We apply the product rule: \( \frac{d}{dx}(uvw) = \left(\frac{du}{dx}vw + u\frac{dv}{dx}w + uv\frac{dw}{dx}\right) \).

Step 3: Differentiate Each Part Using the Chain Rule

1. Differentiate \( u = \tan x \): \( \frac{du}{dx} = \sec^2 x \).2. Differentiate \( v = \tan 2x \): \( \frac{dv}{dx} = 2 \sec^2 2x \), applying the chain rule.3. Differentiate \( w = \tan 3x \): \( \frac{dw}{dx} = 3 \sec^2 3x \), using the chain rule.

Step 4: Substitute into Product Rule Expression

Substitute the derivatives back into the product rule expression: \( \frac{dy}{dx} = (\sec^2 x \cdot \tan 2x \cdot \tan 3x) + (\tan x \cdot 2\sec^2 2x \cdot \tan 3x) + (\tan x \cdot \tan 2x \cdot 3\sec^2 3x) \).

Step 5: Simplify the Expression

Combine the terms: \( \frac{dy}{dx} = \sec^2 x \tan 2x \tan 3x + 2 \sec^2 2x \tan x \tan 3x + 3 \sec^2 3x \tan x \tan 2x \).

Step 6: Identify the Correct Answer

Compare the simplified derivative with the given options. Our expression matches option (a):\( 3 \sec ^{2} 3x \tan x \tan 2x + \sec ^{2} x \tan 2x \tan 3x + 2 \sec ^{2} 2x \tan 3x \tan x \).

Key Concepts

Product Rule

When dealing with functions that are products of other functions, like our function \(y = \tan x \tan 2x \tan 3x\), we must use the product rule for differentiation. The product rule is essential for finding the derivative of products and can be expressed as:

• For two functions, \(u(x)\) and \(v(x)\), the derivative \(\frac{d}{dx}(u \cdot v)\) is \(\frac{du}{dx} \cdot v + u \cdot \frac{dv}{dx}\).
• In our function's case as a product of three functions, say \(u\), \(v\), and \(w\), the derivative is \(\frac{dy}{dx} = \frac{du}{dx} \cdot v \cdot w + u \cdot \frac{dv}{dx} \cdot w + u \cdot v \cdot \frac{dw}{dx}\).

Using this rule helps to methodically handle complex expressions without making mistakes. We first differentiate each part separately before recombining them.

Chain Rule

The chain rule is crucial when you have composite functions, such as \(\tan 2x\) and \(\tan 3x\) in our function \(y = \tan x \tan 2x \tan 3x\). This rule allows us to differentiate the outer function with respect to the inner function and then multiply it by the derivative of the inner function. Here’s how it works:

• For a function \(f(g(x))\), the chain rule tells us that the derivative is \(f'(g(x)) \cdot g'(x)\).
• In our exercise, for the function \(\tan 2x\), we identify \(g(x) = 2x\) and differenitate the outer \(\tan\) as \(\sec^2(2x)\) and multiply it by \(\frac{d}{dx}(2x) = 2\), giving us \(2 \sec^2(2x)\).
• Similarly, for \(\tan 3x\), we go through the same steps to get \(3 \sec^2(3x)\).

The chain rule breaks down complex derivatives into more manageable pieces, making it easier to solve problems step by step.

Trigonometric Functions

Trigonometric functions like tangent, sine, and cosine often appear in differential calculus, particularly in problems involving periodic functions or oscillations. Here's a brief overview relevant to our exercise:

• The tangent function is denoted by \(\tan(x)\), which is defined as the ratio of \(\sin(x)\) to \(\cos(x)\).
• The derivative of \(\tan(x)\) with respect to \(x\) is \(\sec^2(x)\). This derivative is used directly from trigonometric identities during differentiation tasks.
• The secant function, \(\sec(x)\), is the reciprocal of the cosine function, \(1/\cos(x)\).
• We use these trigonometric identities and derivatives to transform and simplify our expressions, ensuring accuracy in each step of the differentiation process.

Comprehending how to handle trigonometric functions in differentiation is fundamental for solving advanced calculus problems accurately.

Derivative Calculation

Calculating derivatives involves systematically applying rules and principles to break down a function into its components. In our example, we've focused on these steps:

• Start by listing components - Separate your expression into parts. For \(y = \tan x \tan 2x \tan 3x\), these parts are \(\tan x\), \(\tan 2x\), and \(\tan 3x\).
• Apply the product rule - As explained, use this rule for the multiple components of your expression.
• Differentiation with the chain rule - Apply this rule to handle functions like \(\tan 2x\) and \(\tan 3x\) by identifying inner functions and multiplying derivatives accordingly.
• Substitute and simplify - Replace derivatives and recombine using algebraic methods to obtain a simplified expression that matches one of the given options.

The act of calculating derivatives integrates understanding of rules and creative algebraic manipulation to express functions in straightforward forms. Mastery of this helps solve a wide variety of calculus problems efficiently.